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I was solving a physics problem and I got $$\frac{a^2+b^2}{2ab}$$

or equivalents

$$\frac{1+ \left( \frac{b}{a}\right)^2}{2\cdot\frac{b}{a}}.$$

If $x=\frac{b}{a}$, we obtain

$$\frac{1+x^2}{2x}$$ or $$\frac{1}{2x}+\frac{x}{2}.$$

I need to know if this expresions are $>1$ when $a<b$ (or $x>1$). I know the answer is yes (I used geogebra to plot) but I'd like to know if there is any analytical way to get the answer. Thank you.

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    $\begingroup$ en.wikipedia.org/wiki/AM_GM_inequality $\endgroup$
    – user296602
    Aug 10, 2017 at 3:54
  • $\begingroup$ If you assume $\,a,b \gt 0\,$ then the question should state so. $\endgroup$
    – dxiv
    Aug 10, 2017 at 3:57
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    $\begingroup$ Presumably by $1/2x$ you mean $\frac 1{2x}$ but if it is read left to right it evaluates as $\frac x2$. Parentheses or stacked fractions, please. $\endgroup$ Aug 10, 2017 at 4:04
  • $\begingroup$ That's equivalent to showing if $\frac 1x + x > 2$. It's well known that if $x \ne 1$ (and $x$ positive that $\frac 1x + x > 2$ so this it true for all positive values of $x$ except $1$. Consider that $\frac 1x + x > 2$ iff $1 + x^2 > 2x$ iff $x^2 - 2x + 1 = (x - 1)^2 > 0$ iff $x \ne 1$. [This ALL assumes $x > 0$] $\endgroup$
    – fleablood
    Aug 10, 2017 at 6:58

3 Answers 3

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From your expression $\frac{x}{2}+\frac{1}{2x}$ for $x>0$ you can use AM-GM: $$\frac{x}{2}+\frac{1}{2x}\geq2\sqrt{\frac{x}{2}\cdot\frac{1}{2x}}=1.$$ The equality occurs for $x=1$, which says that $1$ is a minimal value.

If $x$ real number without condition $x>0$ then since $\lim\limits_{x\rightarrow0^-}\left(\frac{x}{2}+\frac{1}{2x}\right)=-\infty$,

we see that the minimum of this expression does not exist.

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Note that $$0\leq (a-b)^2 = a^2 - 2ab + b^2$$ and hence $$a^2+b^2 \geq 2ab.$$ Therefore, assuming $ab>0$, we have $$ \frac{a^2+b^2}{2ab} \geq 1.$$

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For $x > 0$ (and $x$ must be greater than zero):

$\frac {1}{2x} + \frac x2 > 1 \iff$

$\frac 1x + x > 2 \iff$

$1 + x^2 > 2x \iff $

$x^2 - 2x + 1 > 0 \iff$

$(x - 1)^2 > 0 \iff$

$x - 1 \ne 0 \iff$

$x \ne 1$.

That's it. If $x = 1$ then $\frac 1{2x}+ \frac x2 = 1$ other wise $\frac 1{2x}+ \frac x2 > 1$

If $x < 0$ well .... $\frac 1{2x} + \frac x2 < 0 < 1$ so duh... but we can use the same argument to show that $\frac 1{2x} +\frac x2 < -1$ for $x < 0; x\ne -1$

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